Counting isotone selfmaps and automorphisms is difficult, in general, and Boolean layer cakes are no exception. Zaguia [50] is an excellent reference. Much activity has centered around crowns and fences; we only cite [20] and its bibliography as a point of entry into the literature.
The only place where isotone selfmaps of layer cakes are mentioned in the literature seems to be [14, section 3,] where it is presented as a well-known fact (and referred to [13]) that the level-preserving (and thus order-preserving) automorphisms of P(i,i+1;2i+1) are exactly the automorphisms induced by permutations of the base set which realizes P(i,i+1;2i+1) - in other words, they are just restrictions of the (lattice) automorphisms of the parent Boolean lattice. We will show that this is the case for almost all layer cakes. These results are contained in Lippert's Masters Thesis [38].
Calling a layer i of 
trivial iff i=0 or i=n,
the exceptional cases are the layer cakes P containimg just one
nontrivial layer i with 1<i<n-1: Here the automorphisms are given
by the permutations of the members of this layer, and since 
for these values of i, there are some automorphisms not induced by any
permutation of the base set in this case (if the only nontrivial layer is 1
or n-1, the required permutation is obvious).
The following lemma provides the key step for all remaining cases:
Proof:
Assume P(i,i+1;n) is realized on X, and 
is an automorphism. For the purpose of this proof, call a subset 
good iff there exists an injective map 
such that for
any 
, we have 
whenever 
and
whenever 
. We will prove
inductively that X itself is good which establishes our claim.
Fix any i+1-set . Define 
by
for any 
. Since
for all i-sets 
contained in C and we have
iff 
, we see that 
has the required
properties and hence, that C is good, providing an induction base.
For the induction step, consider any good set and point 
. Pick an i-set contained in U. Since U is good, we
can't have 
(otherwise
, contradicting ). But 
; so define 
to be the unique
point in 
. Let 
be any further i-set contained in U such that
has cardinality i-1, and use 
to construct
analogously. Now 
and so 
. Consider 
: This set contains i points and is
a subset of 
. Since
, we
conclude that y'=y''. Now for any two i-sets 
contained in U there is some sequence 
of
such sets that 
for 
. Using the
preceding argument sequentially, we conclude that the points y' resp. y''
constructed for D and must coincide. It follows that 
is good with 
given by 
, completing the
proof.
We show next that the situation of having two adjacent nontrivial layers as in 2.4.1 may be manufactured whenever there are any two nontrivial layers:
Proof:
Suppose 0<i<j<n are two consecutive nontrivial layers of P and 
.
We show that 
may be extended to P' obtained from P by adding layer
j-1. Let P be realized on X and consider 
, |B|=j-1.
Define 
.
Since is an automorphism, it is injective and we have 
whenever |A|=i. Thus
is a set containing at least 
different subsets
of size i, so certainly 
. Consider 
with
|C|=j and 
. Since preserves order, we see that
and so 
. It
follows that has either j-1 or j elements.
We exclude the latter possibility: If 
, then 
,
so define an j-set 
by 
. Now B is a
proper subset of 
, so select a different j-set 
with
. But then 
,
implying 
, a contradiction. We conclude that 
maps
(j-1)-sets to (j-1)-sets.
Moreover, is injective: Let 
and 
be two different
(j-1)-sets. Then there is a i-set 
such that, say, 
but 
. If now 
, then
. By the preceding part of the proof, we have
that 
, so this set has exactly subsets of
size i which all occur as -images of i-subsets of by
construction. Since is injective, none of them can serve as
-image of A, a contradiction.
Put 
for any 
, then is obviously
order-preserving and bijective, thus an automorphism of P' which completes
the proof.
Proof:
Using Lemma 2.4.2 if necessary we may assume that
P contains a suborder of the form P(i,i+1;n) for some i
with 0<i<i+1<n=|X|. The restriction of any automorphism of
P to P(i,i+1;n) must be induced by some permutation f of X by Lemma
2.4.1.
Let , |A|=k. If k<i, write A as the intersection of all i-sets
. Then 
. If k>i+1, write A as the union of all i+1-sets contained in A
and proceed analogously.
Writing Aut(P) for the set of all automorphisms of any order P, and End(P) for the the set of all isotone self-maps of P, Rival and Rutkowski proposed the following conjecture in [45]:
While open in general. the conjecture has been verified for several large classes of ordered sets, e.g., for cycle-free orders (see [39]) or for lattices (see [40]). We may add layer cakes to the list:
Proof:
Let P be any ordered set with p elements, and let k be the size of a
longest
chain in P. Duffus, Rödl, Sands and Woodrow showed in [15] that in
this case 
.
If P is a layer cake with at least two nontrivial layers
realized on a set X with n elements, these layers of
contribute at least 
elements to P, and the size of a
longest chain is at least 2, hence 
. On the other hand, |Aut(P)| = n! by
2.4.3. So 
. Since
, this ratio tends to zero as n gets large.
If P has exactly one nontrivial layer with 
elements,
|Aut(P)|=k! while 
, so the ratio tends also to zero.
